Chapter 3 Problems 3.7 The following table lists temperatures and specific volumes of water vapor at two pressures: P=1.0 MPa P=1.5 MPa T(°C) T(°C) v(m /kg) /kg) T(°C) T(°C) v(m /kg) /kg) 200 0.2060 200 0.1325 240 0.2275 240 0.1483 280 0.2480 280 Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate: 3
a) The specific volume at T=240°C, P=1.25MPa, (in m /kg). 3 b) The temperature at P=1.5MPa, v=0.1555m /kg, (in °C). 3 c) The specific volume at T=220°C, P=1.4MPa, (in m /kg).
a) At a temperature of 240°C, the specified pressure of 1.25 MPa falls between the table values of 1.0 and 1.5 MPa. To determine the specific volume corresponding to 1.25 MPa, we find the slope of a straight line joining the adjacent table states, as follows:
Similar triangles:
slope v
v 0.1483
0.2275 0.1483
1.5 1. 1.25
1.5 1. 1.0
0.1879
m3 kg
(a )
v
0.1483
0.25 0.50
0.2275 0.1483
3
b) At a pressure of 1.5MPa, the given specific volume of 0.1555m /kg falls between the table values of 240 and 280°C. To determine the temperature corresponding corresponding to the given specific volume, we find the slope of a straight line joining the adjacent table states, as follows:
slope T
T 240
280 240
.1555 .1483 .1627 .1483 .1555 .1483 240 40 .1627 .1483
260 C
(b)
c) In this case, the specified pressure falls between the table values of 1.0 and 1.5MPa and the specified temperature falls between the table values of 200 and 240°C. Thus, double interpolation is required. At 220°C, the specific volume at each pressure is simply the average over the interval: .2060 .2275 m3 At 1.0 MPa, 220°C; v 0.21675 2 kg At 1.5 MPa, 220°C; v
.1325 .1483
m3
0.1404 2 kg Thus with the same approach as in (a) v 0.1404 0.21675 0.1404 0.1 v 0.1404 1.5 1.4 1.5 1.0 0.5
0.15567
m3 kg
(c )
0.21675 0.1404
3.8
The following data lists the temperature and specific volume of NH3 at two pressures: P
50 Lbf/in 2
P 60 Lbf/in 2
T ( F)
v(ft 3 /Lb)
T
v(ft 3/Lb)
100
6.836
100
5.659
120
7.110
120
5.891
140
7.380
140
6.120
a) v
? at T
120 F, P 54Lbf/in2
60 54
v
5.891
v
6.622 ft 3 /Lb
b) T
60 50
? at P
T 120
(7.110 5.891)
60Lbf/in 2 and v
5.982 5.891
6.12 5.891 T 127.9°F c) v
? at T
5.982 f t 3/Lb
(20)
110°F, P 58Lbf/in2
Double interpoloation At 110 F, the specified volume at each pressure is simply the average over the interval: at 50 at 60
lbf 2
, 110°F; v
2
, 110 F; v
in lbf
7.110 6.836
6.973ft3 /lb
2 5.891 5.659
in 2 With the same approach as in (a) v - 5.775
6.973 5.775
60 - 58
60 50
v
5.775ft3 /lb
5.775
2 10
[6.973 5.775]
6.015ft3 /lb
3.10 For H2O, determine the specified property at the indicated state. Locate the state on a sketch of the T-v diagram. 3
a) P=300 kPa, v=0.5 m /kg. Find T, in °C.
Table A-3, v f Since v f
v
1.0732 /103 m3 / kg , v g
0.6058m3 / kg
vg , the state is in the two-phase,
liquid-vapor region, as shown on the T-v diagram. Therefore, T
T sat 3bar
133.6 C
T
3
b) P=28 MPa, T=200°C. Find v, in m /kg.
State is in liquid region. From Table A-5, @200°C 250 bar: v 1.1344 *10 3 300 bar: v 1.1302 *10 So, at 200 bar 3 3 m v 1.1319 *10 kg
3
3
c) P=1 MPa, T=405°C. Find v, in m /kg. From superheat Table A-4 at 10 bar, interpolation gives 3
v = 0.309m /kg.
3
d) T=100°C, x=60%. Find v, in m /kg.
Two phase equation:
v
1 x v f
xvg
With data from Table A-2 at 100°C, v
0.4 1.0435 *10
3
0.6 1.673
1.0042m 3 / kg
3.13
3.18 Determine the qualityof a two phase liquid vapor mixture of: a) H 2O @ P 10 v
(1 x)v f
15 x
Lbf
with
in 2
v 15
ft 3 Lbm
xvg
(1 x)0.01659 x (38.42) 0.39
b) R134a @ T u f
60 °F
30.39 and u g
u
(1 x)u f
x
0.284
91.22
h
(1 x) h f
x
0.486
and
(1 x)v f x
0.246
Btu Lbm
101.27
hg
Lbf
with
in 2 623.32
h=350
Btu Lbm
xhg
d) Pr opane @ T v
u=50.5
xu g
c) Ammonia @ P 80 h f
with
xvg
20°F
with
v 1
ft 3 Lbm
3.21 As shown in Fig. P3.21, a closed, rigid cylinder contains different volumes of saturated liquid water and saturated water vapor at a temperature of 150°C. Determine the quality of the mixture, expressed as a percent. Fig P3.21
Analysis: mvap x , m V / v. Thus, mvap mvap mliq x
Vvap
Vvap vg
, mliq
Vvap / vg Vvap / vg
Vliq / v f
30 A and Vliq
20 A , where area A is in the
same units as the vertical measure shown. Then
30 A / v g
x
30 A / v g
20 A / v f
1 1
20 vg 30 v f
Since ratios appear in the last expression, the quantities can be in any consistent units. Using vf and vg from Table A-2 at 150°C, v f 1.0905 10 3 m3 / kg
v g x
0.3928m3 / kg
1 20 0.3928 1 30 1.0905 10
0.0041 0.41% 3
V liq vf
3.24 Water is contained in a closed tank initially saturated vapor at 200 C is cooled to 100 C. Determine the initial and the final pressures each in bar. Sketch the T-v and the P-v diagram. v1
0.1274
m3 kg
P 1 15.54Bar x1
1
mass & volume does not change (closed system) v1
v2
Since v f P2
v2
vg , the state is in the two-phase,
P sat at 100C 1.014Bar
3.25
3.29 Ammonia contained in a piston cylinder arrangement initially atT 1
0 F , saturated vapor undergoes an isothermal process during which its vo lume (a) doubles, (b) reduces to half. For each case find the final state giving quality or pressure as appropriate. S ketch the process on a T-v and p-v diagram. T 1 v1
0 F v g
v2
2v1
P 2
15.6
T 2
0 F
v2
1
9.11
ft 3 Lb
1.822 Lbf in 2
ft 3 Lb interpolate table 15E
ft 3
v1 4.555 2 Lb Since at T2 0 F v f x
4.555 0.02419 9.11 0.02419
v2 0.499
vg State 2 is two phase 0.5
3.32 Two Lb of water vapor in a piston-cylinder is compressed at constant P 1 250
Lbf in 2
from a volume
of 6.88 ft 3 to saturated vapor state. Determine the temperatue at initial and final state and the work fo r this process in Btu. Solution: 6.88
v1
2
3.44
ft 3 Lb
Knowing v1 and P1 T1
250
Lbf in 2
, The state is sup erheated ( A 4E )
1000 F
v2
v g at P2 250 T2
V2
1.845
in 2
ft 3 Lb
401 F (2 Lb)(1.845
W = PdV W
Lbf
ft 3 Lb
3.69 ft 3
)
P (V2 V1
250(3.69 6.88)
144
147.6 Btu 778 Work done on the system (compression)
3.40 Determine the values of the specified properties at each of the following conditions. 2 a) For Refrigerant 134a at P=140 lbf/in and h=100 Btu/lb, determine T in °F and v in 3 ft /lb. 3 2 b) For ammonia at T = 0°F and v =15 ft /lb, determine P in lbf/in and h in Btu/lb. 3 2 c) For Refrigerant 22 at T=30°F and v =1.2 ft /lb, determine P in lbf/in and h in Btu/lb. 2
a) Refrigerant 134a; p=140lbf/in , h=100 Btu/lb from Table A-11E; hf
(1 x ) h f
xhg 100
x
0.788
v
(1 x ) v f
v
0.2675 ft 3 / lb
T
T sat
(1
100.56 F
0 F , v 15 ft 3 / lb
from Table A-13E since v g v g
x(114.95)
xvg
b) Ammonia; T v
x)44.43
9.1100 ft 3 / lb
superheated vapor
Interpolating in Table A-15E: p
18.85
lbf in
2
and h
c) Refrigerant 22; T from Table A-7E, v g v
v g
615.2
Btu lb
30 F , v 1.2 ft 3 / lb
0.7804 ft 3 / lb
superheated vapor
Interpolating in Table A-9E: lbf Btu p 47.60 2 and h 108.80 in lb
3.52
Solution :
u1
113.83
Q W 1.2 u2
Btu
Lbm m(u2 u1 )
(from table A-12E)
4.32 0.5(u2 113.83) 107.59
Btu
Lbm Knowing P2 & u 2
T2
80 F
3.56 As shown in Fig. P3.56, 0.1 kg of propane is contained within a piston-cylinder assembly at a constant pressure of 0.2 MPa. Energy transfer by heat occurs slowly to the propane, 3 3 and the volume of the propane increase from 0.0277m to 0.0307m . Friction between the piston and cylinder is negligible. The local atmospheric pressure and acceleration of 2 gravity are 100 kPa and 9.81 m/s , respectively. The propane experiences no significant kinetic and potential energy effects. For the propane, determine (a) the initial and final temperature, in °C, (b) the work, in kJ, and (c) the heat transfer, in kJ. Schematic & Given Data:
Engr. Model: 1. The propane within the piston-cylinder assembly is the close system. 2. Friction between the piston and cylinder is negligible and the expansion occurs slowly at a constant pressure of 0.2 MPa. 3. Volume change is the only work mode. 4. For the system there are no significant kinetic and potential energy effects. Analysis: a) To find T1 and T2 requires two property values to fix each state. Since pressure is constant, it is one of the properties. The specific volume provides the other: V 2 0.0307 m3 V 1 0.0277 m3 m3 m3 v1 0.277 0.307 , v2 . Thus, from Table Am 0.1kg kg m 0.1kg kg 18 at 0.2MPa=2bar, T1=30°C, T2=60°C. b) Since volume change is the only work mode and pressure is constant, 2
W12
pdV
p V2 V1
1
0.2MPa 0.0307 0.0277 m3 c) An energy balance reduces to give,
Q12
106 N / m2
1kJ
1 MPa
103 N m
U W12
KE
PE Q12 W 12 or
m u2 u1
With data from Table A-18 at 0.2MPa=2bar
Q12 5.4
0.6kJ Q12
0.6kJ
0.1kg 524.3 476.3
kJ kg
W12
3.77 A system consisting of 1kg H2O undergoes a power cycle composed of the following processes: Process 1-2: Constant-pressure heating at 10 bar from saturated vapor. Process 2-3: Constant-volume cooling to P3= 5bar, T3= 160°C. Process 3-4: Isothermal compression with Q34 = -815.8kJ. Process 4-1: Constant-volume heating. Sketch the cycle on T-v and p-v diagrams. Neglecting kinetic and potential energy effects, determine the thermal efficiency.
Analysis: The thermal efficiency of a power cycle is ή = Wcycle/Qin where Wcycle=W12+W23+W34+W41. And Qin= Q12 Q41 Process 1-2: State 1:
P1=10 bar and x = 1. Therefore v1 = 0.1944 and u1=2583.6
State 2 is fixed by P2=10 bar, but need another property From Table A-4 at P3= 5 bar, T3= 160°C by interpolation between T=151.86 and T=180 m3 KJ v3 0.3835 and u3 2575.2 kg Kg v2
v3
0.3835
m3 kg
and
P2 =10 bar, therefore u2
2
W12
pdV
mp v2
v1
1kg 10bar
105 N / m 2 1bar
1
The first law for closed system is:
Q12
m u2
u1
W12
Q
U
3231.8
0.3835 0.1944
KJ Kg
m3
1kJ
kg 103 N m
W
1kg 3231.8 2583.6
189.1kJ
837.3kJ
189.1kJ
Process 2-3: W 23
0
Q23
m u3
Process 3-4: Q34
State 4 is fixed by T4
W34
w34
815.8
x4ug
1kg 2575.2 3231.8
W34
656.6kJ
Q34
U
Q34 m u4 u3
v1
v4
v f
0.1944 1.102 /10 3
v g
v4
0.3071 1.1021/103
(1 0.6317)674.86
1 1871 2575.2
Process 4-1: W 41 Q41
U
160 C , v4
x4
(1 x4 )u f
0
w23
815.8kJ (given).
Q34
u4
u2
0.6317
0.6317 2568.4
1871kJ / kg
111.6kJ
0 , and U
W
0
1
m u1
u4
1kg 2583.6 1871
kJ
712.6kJ
kg
The net work is then
Wcycle Wcycle
W12 W23 W34 W41
189.1 0
111.6
0
77.5kJ
To obtain Qin,
Qin
Q12
Q41
837.3 712.6 1549.9kJ
Then
77.5 1349.9
0.05 (5%)
As a check, note that for every cycle Qcycle= Wcycle. Qcycle Q12 Q23 Q34 Q41 837.3
656.6
815.8
712.6
77.5kJ
Which agrees with Wcycle calculated using the work quantities.
3.85 A gallon of milk at 68°F is placed in a refrigerator. If energy is removed from the milk by heat transfer at a constant rate of 0.08 Btu/s, how long would it take, in minutes, for the milk to cool to 40°F? The specific heat and density of the milk are 0.94 Btu/lb*°R 3 and 64 lb/ft , respectively. Schematic & Given Data:
Engr. Model:1. The milk is the closed system. 2. For the system, W 0 and kinetic and potential energy play no role. 3. The milk is modeled as incompressible with constant specific heat, C. Analysis: An energy rate balance reduces as follows,
dv dt In this expression, m
u2 u1
t
V
Q
U 64
lb ft 3
Q dt 1gal
dv
dKE
dPE
dt
dt
dt
m u2 u1 0.13368 ft 3 1gal
Q t. 8.56lb . Also, with Eq. 3.20a
c T2 T1 . Collecting results
mc T2 T1 Q
8.56lb
0.94
Btu
lb R 0.08 Btu / s
28 R
1min 60 s
Q W or
47min
3.98 2
Five lbmol of carbon dioxide (CO2), initially at 320 lbf/in , 660R, is compressed at constant pressure in a piston-cylinder assembly. For the gas, W=-2000Btu. Determine the final temperature, in R. Engr. Model: 1. The CO2 is a closed system. 2. Pressure remains constant during the compression process. 2
Analysis: The final state of the gas is fixed by pre ssure, 320 lbf/in , and the specific volume v2 , which can be evaluated from the work. 2
W
pdV
p V2 V1
np v2
v1
v2
v1
1
W np
(*)
v1 can be found from the compressibility chart. From Table A-1E, pc=72.9 atm, Tc=548R. Then 320/14.7 atm 660 R p R 0.30 T R1 1.2 72.9atm 548 R Using these, Fig A-1 gives v R' 1 4.0 , Then
v1
' R1
v
ft lbf 548 R lbmol R 72.9 14.7 lbf / in2
1545
RT c
4.0
pc
1 ft 2 2
144in
21.95 ft 3 / lbmol
Returning to Eq. (*)
v2
W
v1
( 2000 Btu)
3
21.95
np
ft
lbmol
5lbmol
778 ft lbf 1Btu
320 14.8lbf / ft 2
15.21 ft 3 / lbmol So, at state 2, p R 2 V R' 2
pR1
0.30 and v R' 2
v2 pc / RTc
15.21 ft 3 / lbmol 72.9 14.7 144 lbf / ft 2 1545
Returning to Fig. A-1, T R 2
ft lbf lbmol
R
548 R
0.95 . Thus T2
T R 2TC
0.95 548 R
521 R
2.77
3.128 Air is confined to one side of a rigid container divided by a partition, as shown in Fig. P3.128. The other side is initially evacuated. The air is initially at p1=5 bar, T1=500 K, 3 and V1=0.2m . When the partition is removed, the air expands to fill the entire chamber. Measurements show that V2=2V1 and p2=p1/4. Assuming the air behaves as an ideal gas, determine (a) the final temperature, in K, and (b) the heat transfer, kJ. P3.123
Engr. Model: 1. The closed system is the region within the container, ignoring the partition. 2. The air is modeled as an ideal gas. 3. There are no overall changes in kinetic or potential energy. 4. W=0.
Analysis: a) Using the ideal gas model equation of state, PV 1 1 T2
T1
PV 2 2 PV 1 1
b) An energy balance reduces to Q
500 K
1
Q
m u2
2
4
W U
mRT 1 , PV 2 2
KE
250 K
mRT 2 . Thus,
PE , or
u1
Where m
PV 1 1
5 105 N / m2
RT 1
8314 N m 28.97 kg K
0.2m2
0.7kg
500 K
So, with data from the air Table A-22 Q
0.7kg 178.28 359.49
kJ kg
126.8kJ
3.135 Lbf
, T1 500R to a final in 2 3 volume of V2 = 1 ft in a process described by pv1.25 const . The mass of the air is 0.5 Lb. Assuming ideal gas behavior, and ignoring kinetic and potential energy effects, determine the work and heat transfer, each in Btu. Solve the problem each of two ways: (a) Using a constant specified heat evaluated at 500 R. (b) Using data from Table A-22E. Compare the results and discuss Air is compressed in a piston cylinder assembly from p1
Rair v2 v1
M air 1
2
0.5
T2
ft.Lbf
) Lbm.R 28.97
R
53.3
ft 3 Lbm
RT
(53.3)(500)
p1
(10)(144)
p1v11.25 p2
1545(
18.52
ft 3 Lbm
p2 v21.25
161.5
Lbf
p2v2
in 2 (161.5)(144)(2)
R
(53.3)
W
(0.5)
Q
W
p2 v2
U2
p1v1 1
1 n U1
872 0 R 51 Btu
778
Using the ideal gas model: u2 Q
u1
Cvavg (T2 T1 ) 0.173(872 500)
51 0.5(64.356)
18.82 Btu
Using the Air Table A22E u1 =85.2
Btu
Lbm Btu u 2 =149.61 Lbm Q 51 0.5(149.61 85.2)
18.80 Btu
64.356
Btu Lbm
10
3.138 Two-tenths kmol of nitrogen (N2) in a piston-cylinder assembly undergoes two processes in series as follows: 3 3 Process 1-2: Constant pressure at 5 bar from V1=1.33m to V2=1m . Process 2-3: Constant volume to p3=4bar. Assuming ideal gas behavior and neglecting kinet ic and potential energy effects, determine the work and heat transfer for each process, in kJ. Schematic & Given Data: Engr. Model: 1) The N2 contained in the piston-cylinder assembly if the closed system. 2) Volume change is the only work mode. 3) The N2 is modeled as an ideal gas. 4) Kinetic and potential energy effects play no role. Analysis: Process 1-2 occurs at constant pressure. Thus 2
W12
1
W12
pdV
p V2 V1 . That is:
5bar 1m
3
1.33m
3
105 N / m2
1kJ
1bar
103 N m
165kJ
Process 2-3 occurs at constant volume. That is W23=0. Reducing an energy balance, V KE PE Q W . Thus Q requires T1,T2,T3. To find T1 was the ideal gas equation of state: N 5 105 2 1.33m3 PV m 1 1 T1 400K 8314 N m nR 0.2kmol Kmol K
W
For process 1-2, p=constant; so the ideal ga s equation of state gives pV1
pV2
n u . This
nRT 1 ,
nRT 2 , or T2
T1
V 2 V1
For process 2-3, V=constant; so p2V T3
P 3 P2
T2
400
1m3 1.33m3
nRT 2 , p3V 4bar 5bar
301K
nRT 3 or
301K
241K
For process 1-2, with u1 and u 2 from Table A-23
Q12
W12
n u2
u1
For process 2-3 Q23
165kJ W23
0.2kmol 6250 8314
n u3 u2
kJ kmol
0.2kmol 5000 6250
578kJ kJ kmol
250kJ
3.140 Air is contained in a piston cylinder assembly un dergoes the power cycle shown. Assuming ideal gas behavior, evaluate the thermal efficiency of the cycle.
3.140 continued: Analysis: w12 w12 q12
2 1
pdv
1*105 3
10 w12
c ln ln
5 1
v2
Pv 1 1 ln
v1
804.7
v2 v1
KJ Kg
u12
Since no change in temperature (isothermal) q12 w23 w23 T2 T3
804.7 2 1
Kg
1*105 3
10 R P3v3 R
p v3
v2
(1 5)
400
w31
0 Volume is const .
wnet
404.7
wnet qin
Kg
348.4 K from air table A22
w23
q12
KJ
1742.2 K from air table A22
q23
qin
0
KJ
pdv
P2v2
u12
u3
u2
KJ Kg
q23
1583.6
and
1988.3
qnet KJ Kg
0.204 (20.4%)
KJ Kg
404.7
KJ Kg
u2 u3
1432.5 248.9
KJ Kg
KJ Kg
